Questions

Q 1

MS 5.54 - pg 215 Phishing attacks to email accounts. Refer to the Chance (Summer, 2007) article on phishing attacks at a company, Exercise 2.24 (p. 38). Recall that phishing describes an attempt to extract personal/financial information through fraudulent email. The company set up a publicized email account—called a “fraud box”—which enabled employees to notify them if they suspected an email phishing attack. If there is minimal or no collaboration or collusion from within the company, the interarrival times (i.e., the time between successive email notifications, in seconds) have an approximate exponential distribution with a mean of 95 seconds.

  1. What is the probability of observing an interarrival time of at least 2 minutes? \[\int_{120}^{\infty}\frac{e^{-y/95}}{95} dy = -e^{-y/95}\Big|_{120}^{\infty} = e^{-\infty/95}+e^{-120/95}\]
0+exp(-120/95)
## [1] 0.2827597
  1. Data for a sample of 267 interarrival times are saved in the PHISHING file. Do the data appear to follow an exponential distribution with \(\beta\) = 95?
phishing <- read.csv("PHISHING.csv")
hist(phishing$INTTIME, xlab = "INTTIME", main = "Histogram or phishing")

Yes, this distribution looks like a exponential distribution due to the shape.

Q 2

MS 5.56 - pg 215 Flood level analysis. Researchers have discovered that the maximum flood level (in millions of cubic feet per sec- ond) over a 4-year period for the Susquehanna River at Harrisburg, Pennsylvania, follows approximately a gamma distribution with \(\alpha\) = 3 and \(\beta\) = .07 (Journal of Quality Technology, Jan. 1986).

  1. Find the mean and variance of the maximum flood level over a 4-year period for the Susquehanna River.
alpha = 3
beta = 0.07
mean = alpha*beta
var = alpha*(beta)^2
mean
## [1] 0.21
var
## [1] 0.0147
  1. The researchers arrived at their conclusions about the maximum flood level distribution by observing maxi- mum flood levels over 4-year periods, beginning in 1890. Suppose that over the next 4-year period the maximum flood level was observed to be .60 million cubic feet per second. Would you expect to observe a value this high from a gamma distribution with \(\alpha\) = 3 and \(\beta\) = .07? What can you infer about the maximum flood level distribution for the 4-year period observed?
# High end
mean+3*sqrt(var)
## [1] 0.5737307
# Low end
mean-3*sqrt(var)
## [1] -0.1537307

This event could happen, but it would be extremely rare, definitely an outlier case on the high end.

Q 3

MS 5.60 - pg 216 Reaction to tear gas. The length of time Y (in minutes) re- quired to generate a human reaction to tear gas formula A has a gamma distribution with \(\alpha\) = 2 and \(\beta\) = 2. The dis- tribution for formula B is also gamma, but with \(\alpha\) = 1 and \(\beta\) = 4.

  1. Find the mean length of time required to generate a human reaction to tear gas formula A. Find the mean for formula B.
alpha1 = 2
beta1 = 2
alpha2 = 1
beta2 = 4
mean1 = alpha1*beta1
mean2 = alpha2*beta2
mean1
## [1] 4
mean2
## [1] 4
  1. Find the variances for both distributions.
var1 = alpha1*(beta1)^2
var2 = alpha2*(beta2)^2
var1
## [1] 8
var2
## [1] 16
  1. Which tear gas has a higher probability of generating a human reaction in less than 1 minute? (Hint: You may use the fact that \(\int ye^{-y/2} dy = -2ye^{-y/2} + \int 2e^{-y/2} dy\) This result is derived by integration by parts.)

For A:\[P(Y<1) = \int_{0}^{1} \frac{y^{2-1}e^{-y/2}}{2^2\Gamma(2)} dy = \int_{0}^{1}\frac{ye^{-y/2}}{4} dy\] \[= \frac{1}{4}(-2ye^{-y/2})\Big|_{0}^{1} + \frac{1}{4}\int_{0}^{1}2e^{-y/2} dy\] \[= \frac{1}{4}(-2e^{-1/2} + 2(0)e^0)+(-e^{-y/2})\Big|_{0}^{1} = -0.3033 -e^{-1/2}+e^0\] \[-0.3033-0.6065+1=0.0902\] For B:\[P(Y<1) = \int_{0}^{1} \frac{y^{1-1}e^{-y/4}}{4^1\Gamma(1)} dy = \int_{0}^{1}\frac{e^{-y/4}}{4} dy\] \[= -e^{-y/4}\Big|_{0}^{1} = -e^{1/4} + e^0\] \[= 1-0.7788 = 0.2212\]

Because 0.2212 > 0.0902. This means that Formula B has a higher probability of generating a human reaction in less than 1 minute

Q 4

MS 5.74 - pg 219 Washing machine repair time. Based on extensive testing, a manufacturer of washing machines believes that the distribution of the time Y (in years) until a major repair is required has a Weibull distribution with \(\alpha\) = 2 and \(\beta\) = 4.

  1. If the manufacturer guarantees all machines against a major repair for 2 years, what proportion of all new washers will have to be repaired under the guarantee? \[F(y) = 1-e^{-y^\alpha/\beta}\]
1-exp(-2^2/4)
## [1] 0.6321206
  1. Find the mean and standard deviation of the length of time until a major repair is required. \[\mu = \beta^{1/\alpha}\Gamma(\frac{\alpha+1}{\alpha})\]
mean = 4^{1/2}*gamma(3/2)
mean
## [1] 1.772454

\[\sigma^2=\beta^{2/\alpha}[\Gamma(\frac{\alpha+2}{\alpha})-\Gamma^2(\frac{\alpha+}{\alpha})]\]

sigmaSquared = 4^(2/2)*((gamma((2+2)/2))-(gamma(3/2))^2)
sigma = sqrt(sigmaSquared)
sigma
## [1] 0.9265028
  1. Find \(P(\mu-2\sigma \leq Y \leq \mu + 2\sigma\).
# High range
high = mean+2*sigma
high
## [1] 3.625459
# Low range
low = mean-2*sigma
low
## [1] -0.08055165
(1-exp(-high^2/4))-(1-exp(-0^2/4))
## [1] 0.9625964
  1. Is it likely that Y will exceed 6 years?

No the Probability is extremely small.

1-(1-exp(-6^2/4))
## [1] 0.0001234098

Q 5

MS 5.84 - pg 223 Laser color printer repairs. The proportion Y of a data- processing company’s yearly hardware repair budget allo- cated to repair its laser color printer has an approximate beta distribution with parameters \(\alpha\) = 2 and \(\beta\) = 9.

  1. Find the mean and variance of Y. \[E(y)=\mu=\frac{\alpha}{\alpha+\beta}\]
mean = (2/(2+9))
mean
## [1] 0.1818182

\[\sigma^2=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\]

var = (2*9)/((2+9)^2*(2+9+1))
var
## [1] 0.01239669
  1. Compute the probability that for any randomly selected year, at least 40% of the hardware repair budget is used to repair the laser color printer. \[1-\frac{1}{B(2,9)}\int_{0}^{0.4} y(1-y)^8dy\]
1-1/beta(2,9)*(-0.4*(1-0.4)^9/9-(((1-0.4)^10/(9*10))-(1)^10/(9*10)))
## [1] 0.0463574
  1. What is the probability that at most 10% of the yearly repair budget is used for the laser color printer? \[\frac{1}{B(2,9)}\int_{0}^{0.1} y(1-y)^8dy\]
1/beta(2,9)*(-0.1*(1-0.1)^9/9-(((1-0.1)^10/(9*10))-(1)^10/(9*10)))
## [1] 0.2639011

Q 6

MS 5.114 - pg 232 Lifelengths of memory chips. The lifelength Y (in years) of a memory chip in a laptop computer is a Weibull random variable with probability density \[ f(y) = \begin{cases} \frac{1}{8}ye^{-y^2/16} \quad \text{if } 0\leq y < \infty\\ 0 \quad \quad \quad \quad \quad \quad \text{elsewhere} \end{cases} \]

  1. What are the values of \(\alpha\) and \(\beta\)?
alpha = 2
beta = 16
alpha
## [1] 2
beta
## [1] 16
  1. Compute the mean and variance of Y.
mean = beta^{1/alpha}*gamma((alpha+1)/alpha)
var = beta^{2/alpha}*(gamma((alpha+2)/alpha)-(gamma((alpha+1)/alpha))^2)
mean
## [1] 3.544908
var
## [1] 3.433629
  1. Find the probability that a new memory chip will not fail before 6 years.
1-(1-exp(-(6)^alpha/beta))
## [1] 0.1053992

Q 7

MS 6.2 - pg 239 Tossing dice. Consider the experiment of tossing a pair of dice. Let X be the outcome (i.e., the number of dots appearing face up) on the first die and let Y be the outcome on the second die.

  1. Find the joint probability distribution p(x, y). \(1 \leq X \leq 6\) and \(1 \leq Y \leq 6\).
(1/6)*(1/6)
## [1] 0.02777778

\[p(x, y) = \frac{1}{36}\]

  1. Find the marginal probability distributions \(p_1(x)\) and \(p_2(y)\). \[p_1(x) = P(X=1) = p(1,1)+p(1,2)+p(1,3)+p(1,4)+p(1,5)+p(1,6)\]

\[\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}= \frac{6}{36} = \frac{1}{6}\]

\[p_2(x) = P(X=2) = p(2,1)+p(2,2)+p(2,3)+p(2,4)+p(2,5)+p(2,6)\] \[\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}= \frac{6}{36} = \frac{1}{6}\]

  1. Find the conditional probability distributions \(p_1(x | y)\) and \(p_2(y | x)\).

When y = 1 \(p_2(x | 1)=\frac{p(x,1)}{p_2(1)}\) \[p_1(1|1)=\frac{p(1,1)}{p_2(1)}=\frac{1/36}{1/6}=\frac{1}{6}\] And so on for other \(p_1(x | y)\) … \[\frac{1}{6}\] Meaning \[p_1(x|y) = \frac{1}{6}\] Same probability for \(p_2(y|x)\) because \(1 \leq X \leq 6\) and \(1 \leq Y \leq 6\) (Both sides are equal).

  1. Compare the probability distributions of parts b and c. What phenomenon have you observed?

\(p_1(x) = p_1(x|y)\) which implies that x and y are independent.

Q 8

MS 6.4 - pg 240 Modeling the behavior of granular media. Refer to the Engineering Computations: International Journal for Computer-Aided Engineering and Software (Vol. 30, No. 2, 2013) study of the properties of granular media (e.g., sand, rice, ball bearings, and flour), Exercise 3.62 (p. 120). The study assumes there is a system of N non-interacting granular particles, where the particles are grouped according to energy level, r. For this problem (as in Exercise 3.62), assume that N = 7 and r = 3, then consider the scenario where there is one particle (of the total of 7 particles) at energy level 1, two particles at energy level 2, and four particles at energy level 3. Another feature of the particles studied was the position in time where the particle reached a certain entropy level during compression. All particles reached the desired entropy level at one of three time periods, 1, 2, or 3. Assume the 7 particles had the characteristics shown in the table. Consider a randomly selected particle and let X represent the energy level and Y the time period associated with particle.

knitr::include_graphics("Q8.png")

  1. Find the bivariate probability distribution, p(x, y).
1 2 3
x 1 1/7 2/7 1/7
y 2 0 0 2/7
3 0 0 1/7
  1. Find the marginal distribution, \(p_1(x)\). \[P(X=1)=p_1(1)=p(1,1)+p(1,2)+p(1,3)=\frac{1}{7}+0+0=\frac{1}{7}\] \[P(X=2)=p_1(2)=p(2,1)+p(2,2)+p(2,3)=\frac{2}{7}+0+0=\frac{2}{7}\] \[P(X=3)=p_1(3)=p(3,1)+p(3,2)+p(3,3)=\frac{1}{7}+\frac{2}{7}+\frac{1}{7}=\frac{4}{7}\]
x 1 2 3
\(p_1(x)\) 1/7 2/7 4/7
  1. Find the marginal distribution, \(p_2(x)\). \[P(Y=1)=p_2(1)=p(1,1)+p(2,1)+p(3,1)=\frac{1}{7}+\frac{2}{7}+\frac{1}{7}=\frac{4}{7}\] \[P(Y=2)=p_2(2)=p(1,2)+p(2,2)+p(3,2)=0+\frac{2}{7}+0=\frac{2}{7}\] \[P(Y=3)=p_2(3)=p(1,3)+p(2,3)+p(3,3)=0+0+\frac{1}{7}=\frac{1}{7}\]
y 1 2 3
\(p_2(y)\) 4/7 2/7 1/7
  1. Find the conditional distribution, \(p_2(y | x)\).

When \(X=1\), \(p_2(y|1)=\frac{p(1,y)}{p_1(x)}\) \[p_2(1|1) = \frac{p(1,1)}{p_1(1)} = \frac{1/7}{1/7}=1\] \[p_2(2|1) = \frac{p(1,2)}{p_1(1)} = \frac{0}{1/7}=0\] \[p_2(3|1) = \frac{p(3,1)}{p_1(1)} = \frac{0}{1/7}=0\]

y 1 2 3
\(p_2(y|1)\) 1 0 0

Same reasoning with when \(X=2\) and \(X=3\)

For \(X=2\)

y 1 2 3
\(p_2(y|2)\) 1 0 0

For \(X=3\)

y 1 2 3
\(p_2(y|3)\) 1/4 2/4 1/4

Q 9

MS 6.12 - pg 244 Distribution of low bids. The Department of Transportation (DOT) monitors sealed bids for new road construction. For new access roads in a certain state, let X = low bid (thousands of dollars) and let Y = DOT estimate of fair cost of building the road (thousands of dollars). The joint probability density of X and Y is \[f(x, y) = \frac{e^{-y/10}}{10y},\hspace{8 mm} 0 < y < x < 2y\]

  1. Find f(y), the marginal density function for Y. Do you recognize this distribution? \[f_2(y)=\int_{y}^{2y}f(x,y)dx=\int_{y}^{2y}\frac{e^{-y/10}}{10y}dx\] \[=\frac{e^{-y/10}}{10y}x \Big|_{y}^{2y}=\frac{e^{-y/10}}{10y}(2y-y)=\frac{e^{-y/10}}{10}\] This is an exponential distribution, \(\beta = 10\).

  2. What is the mean DOT estimate, E(Y)?

\(E(Y)=\mu=\beta = 10\).

Q 10

MS 6.14 - pg 245 Servicing an automobile. The joint density of X, the total time (in minutes) between an automobile’s arrival in the service queue and its leaving the system after servicing, and Y, the time (in minutes) the car waits in the queue be- fore being serviced, is \[f(x,y) = \begin{cases} ce^{-x^2} \quad \text{if } 0\leq y < x; 0 \leq x < \infty\\ 0 \quad \quad \quad \text{elsewhere} \end{cases}\]

  1. Find the value of c that makes f(x, y) a probability density function. \[\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)dydx = 1\] \[\int_{0}^{\infty}\int_{0}^{x}ce^{-x^2}dydx=\int_{0}^{\infty}ce^{-x^2}y\Big|_{0}^{x}dx = \int_{0}^{\infty}ce^{-x^2}x dx\] \[\int_{0}^{\infty}cxe^{-x^2}dx=-\frac{cxe^{-x^2}}{2}\Big|_{0}^{\infty}=\frac{c}{2}=1=>c=2\]

  2. Find the marginal density for X and show that \[\int f_1(x)dx = 1\] \[f_1(x)=\int_{-\infty}^{\infty}f(x,y)dy=\int_{0}^{x}2e^{-x^2}dy=2e^{-x^2}y\Big|_{0}^{x}=2xe^{-x^2}\] \[\int_{-\infty}^{\infty}f_1(x)dx=\int_{0}^{\infty}2xe^{-x^2}dx=-e^{-x^2}\Big|_{0}^{\infty}=0+1=1\]

  3. Show that the conditional density for Y given X is a uniform distribution over the interval \(0 \leq Y \leq X\). \[f_2(y|x)=\frac{f(x,y)}{f_1(x)}=\frac{2e^{-x^2}}{2xe^{-x^2}}=\frac{1}{x} \hspace{1cm} 0 \leq Y \leq X\]

Q 11

MS 6.51 - pg 253 As an illustration of why the converse of Theorem 6.6 is not true, consider the joint distribution of two discrete random variables, X and Y, shown in the accompanying table. Show that Cov(X, Y) = 0, but that X and Y are dependent.

knitr::include_graphics("Q11.png")

\[Cov(X,Y)=E(XY) \hspace{1 cm} E(X)E(Y)\] \[E(XY)=\sum_{x}\sum_{y}xyp(x,y)=(-1)(-1)(\frac{1}{12})+(-1)(0)(\frac{2}{12})+(-1)(1)(\frac{1}{12})\] \[+(0)(-1)(\frac{2}{12}+(0)(0)(0)+(0)(1)(\frac{2}{12})+(1)(-1)(\frac{1}{12})+(1)(\frac{2}{12})(0)+(1)(1)(\frac{1}{12})\] \[\frac{1}{12}+0-\frac{1}{12}+0+0+0-\frac{1}{12}+\frac{1}{12}=0\] Then \[P(X=1)=p_1(-1)=p(-1,-1)+p(-1,0)+p(-1,0)=\frac{1}{12}+\frac{2}{12}+\frac{1}{12}=\frac{4}{12}=\frac{1}{3}\] \[P(X=0)=p_1(0)=p(0,-1)+p(0,0)+p(0,0)=\frac{2}{12}+0+\frac{2}{12}=\frac{4}{12}=\frac{1}{3}\] \[P(X=1)=p_1(1)=p(1,-1)+p(1,0)+p(1,0)=\frac{1}{12}+\frac{2}{12}+\frac{1}{12}=\frac{4}{12}=\frac{1}{3}\] \[E(X)=\sum_{x}xp_1(x)=(-1)(\frac{1}{3})+(0)(\frac{1}{3})+(1)(\frac{1}{3})=0\] Same thing for \((Y=-1)\), \((Y=0)\), \((Y=1)\).

\[E(Y)=\sum_{x}yp_1(-1,1)=(-1)(\frac{1}{3})+(0)(\frac{1}{3})+(1)(\frac{1}{3})\]

\[Cov(X,Y)=E(XY)-E(X)E(Y)=0-0(0)=0\] \[p(x,y)=p_1(x)p_2(y)=> \frac{1}{12}\neq(\frac{1}{3})(\frac{1}{3})\] Thus, \(X\) and \(Y\) are not independent even though \(Cov(X,Y) = 0\).

Q 12

MS 6.74 - pg 269 Uranium in the Earth’s crust. Refer to the American Mineralogist (October 2009) study of the evolution of uranium minerals in the Earth’s crust, Exercise 5.17 (p. 199). Recall that researchers estimate that the trace amount of uranium Y in reservoirs follows a uniform distribution ranging between 1 and 3 parts per million. In a random sample of n = 60 reservoirs, let \(\overline{Y}\) represent the sample mean amount of uranium.

  1. Find \(E(\overline{Y})\) and interpret its value. \[E(\bar{Y})=E(\frac{\sum_{i=1}^{60} Y_i}{n})=\frac{1}{n}E(\sum_{i=1}^{60}Y_i)=\frac{1}{60}[60(2)]=2\]

  2. Find Var \(\overline{Y}\). \[\sigma^2_{\bar{Y}}=V(\bar{Y})=V(\frac{\sum_{i=1}^{60} Y_i}{n})=(\frac{1}{n})^2V({\sum_{i=1}^{60} Y_i})\]

sigmaSquared = (1/60^2)*(60/3)
sigmaSquared
## [1] 0.005555556

  1. Describe the shape of the sampling distribution of \(\overline{Y}\). By the central Limit Theorem, the sampling distribution of \(\bar{Y}\) is approximately normal.

  2. Find the probability that \(\overline{Y}\) is between 1.5 ppm and 2.5 ppm.

sigma = sqrt(sigmaSquared)
sigma
## [1] 0.0745356

\[P(1.5 \leq \bar{Y} \leq 2.5)=P(\frac{1.5-2}{0.0745} \leq Z \leq \frac{2.5-2}{0.0745})=P(-6.71 \leq Z \leq 6.71)\] \[=P(0 \leq Z \leq 6.71)+P(0 \leq Z \leq 6.71) \approx 0.5+0.5=1\]

  1. Find the probability that \(\overline{Y}\) exceeds 2.2 ppm. \[P(Y \leq 2.2)=P(Z \geq \frac{2.2-2}{0.0745})=P(Z\geq2.68)\] \[=0.5-P(0 \leq Z \leq 2.68)=0.5+0.4963=0.9963\]

Q 13

MS 6.90 - pg 273 Mercury contamination of swordfish. Consumer Reports found widespread contamination of seafood in New York and Chicago supermarkets. For example, 40% of the swordfish pieces available for sale have a level of mercury above the Food and Drug Administration (FDA) limit. Consider a random sample of 20 swordfish pieces from New York and Chicago supermarkets.

  1. Use the normal approximation to the binomial to calculate the probability that fewer than 2 of the 20 swordfish pieces have mercury levels exceeding the FDA limit. \[P(Y<2)=P(Z < \frac{1.5-8}{2.1909})=P(Z<2.97)=0.5-P(-2.97 < Z <0)=0.5 -0.4985=0.0015\]

  2. Use the normal approximation to the binomial to calculate the probability that more than half of the 20 swordfish pieces have mercury levels exceeding the FDA limit. \[P(Y>10)=P(Z>\frac{10.5-8}{2.1909})=P(Z>1.14)=0.5-P(0<Z<1.14)=0.5-0.3729 = 0.1271\]

  3. Use the binomial tables to calculate the exact probabilities in parts a and b. Does the normal distribution provide a good approximation to the binomial distribution? \[P(Y<2)=P(Y<1)=0.0005\] \[P(Y>10)-1-P(Y \leq 10)-1-0.8725-0.1275\] The normal approximation provides a good estimate of the binomial distribution.

Q 14

MS 7.108 - pg 362 Lead and copper in drinking water. Periodically, the Hillsborough County (Florida) Water Department tests the drinking water of homeowners for contaminants such as lead and copper. The lead and copper levels in water specimens collected for a sample of 10 residents of the Crystal Lake Manors subdivision are shown next.

knitr::include_graphics("Q14.png")

  1. Construct a 99% confidence interval for the mean lead level in water specimens from Crystal Lake Manors.
v = c(1.32,0,13.1,.919,.657,3.0,1.32,4.09,4.45,0)
mu = mean(v)
mu
## [1] 2.8856
sigma = sqrt(var(v))
sigma
## [1] 3.924775
# Range Estimate
mu-sigma
## [1] -1.039175
mu+sigma
## [1] 6.810375

(-1.15, 6.92)

  1. Construct a 99% confidence interval for the mean cop- per level in water specimens from Crystal Lake Manors. (0.1519, 0.6647)

  2. Interpret the intervals, parts a and b, in the words of the problem. We are 99% that the mean lead level in water specimens from Crystal Lake Manors is between -1.5 and 6.92. Since the lead level cannot be negative, we are 99% confident that the mean lead level in water specimens from Crystal Lake Manors is between 0 and 6.92

  3. Discuss the meaning of the phrase, “99% confident.” 99% confidence means that in repeated sampling, 99% of all intervals constructed in a similar manner will contain the true population mean.

Q 15

MS 7.114 - pg 364 Solar irradiation study. The Journal of Environmental Engineering (Feb. 1986) reported on a heat transfer model designed to predict winter heat loss in wastewater treatment clarifiers. The analysis involved a comparison of clear-sky solar irradiation for horizontal surfaces at different sites in the Midwest. The day-long solar irradiation levels (in BTU/sq. ft.) at two midwestern locations of different latitudes (St. Joseph, Missouri, and Iowa Great Lakes) were recorded on each of seven clear-sky winter days. The data are given in the table. Find a 95% confidence interval for the mean difference between the day-long clear-sky solar irradiation levels at the two sites. Interpret the results.

knitr::include_graphics("Q15.png")

For confidence coefficient 0.95, \(\alpha\) = 0.05 and \(\alpha\)/2 = 0.05/2 = 0.025. v = n-1 = 7-1 = 6 degrees of freeedom, \(t_{0.025}\) = 2.447. The 95% confidence interval is: \[\bar{d} \pm t_{\alpha/2}\frac{s_4}{\sqrt{n}} =>198.0 \pm 2.447 \frac{44.5}{\sqrt{7}} => 19.0 \pm 41.16 => (156.84, 239.16)\] We are 95% confident that the true mean difference between the day-long clear-sky solar irradiation levels at the two sites is between 156.84 and 239.16.

Q 16

MS 7.116 - pg 364 Diazinon residue in orchards. Pesticides applied to an extensively grown crop can result in inadvertent areawide air contamination. Environmental Science & Technology (Oct. 1993) reported on air deposition residues of the insecticide diazinon used on dormant orchards in the San Joaquin Valley, California. Ambient air samples were collected and analyzed at an orchard site for each of 11 days during the most intensive period of spraying. The levels of diazinon residue (in mg/m3) during the day and at night are recorded in the table. The researchers want to know whether the mean diazinon residue levels differ from day to night.

knitr::include_graphics("Q16.png")

  1. Analyze the data using a 90% confidence interval. \[\bar{d}=\frac{\sum{d}}{n}=38.909; s_4^2=\frac{\sum{d_1^2}-\frac{(\sum{d_1})^2}{n}}{n-1} = \frac{30,033.96-\frac{(-428)^2}{11}}{11-1} = 1,338.0869\] \[s_4 = \sqrt{1,338.0869} = 36.5799\] Let \(\mu_1\) = mean diazinon residue during the day and \(\mu_2\) = mean diazinon residue at night. Then \(\mu_d = \mu_1-\mu_2\) is the difference between the mean diazinon residue for day and night.

For confidence coefficient 0.90, \(\alpha\) = 0.10 and \(\alpha\)/2 = 0.10/2 = 0.05. v = n-1 = 11-1 = 10 degrees of freedom, \(t_{0.05}=1.812\). The 90% confidence interval is: \[\bar{d} \pm t_{\alpha/2}\frac{s_4}{\sqrt{n}} =>38.9 \pm 1.812 \frac{36.5799}{\sqrt{11}} => -38.909 \pm 19.985 => (-58.894, -18.934)\]We are 95% confident that the true mean difference between the day-long clear-sky solar irradiation levels at the two sites is between -58.894 and -18.924.

  1. What assumptions are necessary for the validity of the interval estimation procedure of part a? We must assume the population of differences is normal.

  2. Use the interval, part a, to answer the researchers’ question. Since the confidence interval in part a does not contain 0, there is evidence of a difference in the mean diazinon residue between day and night. Since the interval contains only negative number, the mean diazinon residue for night is greater than the mean diazinon residue for day.